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Q. In the rectangle, the two corners have charges $q _{1}=-8 \mu C$ and $q _{2}=+4 \mu C$. The work done in moving a charge $+3 \mu C$ from B to A
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Electrostatic Potential and Capacitance

Solution:

Potential at A
$V _{ A }=\frac{ Kq _{1}}{ r }+\frac{ Kq _{2}}{ r }$
$V _{ A }= K \left[\frac{-8 \times 10^{-6}}{20 \times 10^{-2}}+\frac{4 \times 10^{-6}}{5 \times 10^{-2}}\right]$
$V _{ A }= K \times 10^{-4}[-0.4+0.8]$
$V _{ A }= 0 . 4 \times K \times 1 0 ^{-4} V$
Potential at $B V _{ B }=\frac{ K q _{1}}{ r }+\frac{ K q _{1}}{ r }$
$V _{ B }= K \left[\frac{4 \times 10^{-6}}{20 \times 10^{-2}}-\frac{8 \times 10^{-6}}{5 \times 10^{-2}}\right]$
$V _{ B }= K \times 10^{-4}[0.2-1.6]$
$V _{ B }=-1.4 K \times 10^{-4} V$
Charge move from $B$ to $A$ then
$W = q \left( V _{ A }- V _{ B }\right)$
$W =3 \times 10^{-6}\left(0.4 K \times 10^{-4}-\left(-1.4 K \times 10^{-4}\right)\right)$
$\therefore W =3 \times 10^{-6} \times 1.8 \times 9 \times 10^{-4} $
$W =48.6 \times 10^{-1}$
$=4.86 J \approx 4.9 \,J$